Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mirrors traveling near c

  1. Mar 5, 2009 #1
    I've been reviewing time dilation and minkowski diagrams in my spare time and a thought occurred to me which I wanted to confirm. I thought of this when imagining reflections in eyes, so for the sake of simplicity, let's say that mirrors can see.
    Let's say you have two mirrors which are placed face to face at time t = 0. Let's say that mirror B then quickly (for the sake of simplicity: instantly) accelerates to 0.5c (away from mirror A). After 10 seconds at this velocity, mirror B would be seeing mirror A as mirror A was at t = 8.67 seconds. Incidentally, mirror A (at 8.67 seconds) was reflecting the image of mirror B as it was at 7.5 seconds... which was what mirror A was at 6.494 seconds... which was mirror B at 5.62 second... and so on and so forth. After a certain number of reflections, would you get the image of a black mirror (as it was at t=0)? Or would the "instant of reflection" approach t = 0 asymptotically?
     
  2. jcsd
  3. Mar 5, 2009 #2

    Ich

    User Avatar
    Science Advisor

    Yes, but it might be less confusing if you talk about two mirrors approaching each other, with a single photon being reflected between them. Throwing around an "instant of reflection" is not really comprehensible.
    This photon is remarkable: in a finite time, it gets infinitely often reflected, infinitely blueshifted and exerts infinite pressure.
     
  4. Mar 5, 2009 #3
    Ok, I was with you up until this part:
    how can I bring two mirrors together if there's infinite pressure being exerted? I must be misunderstanding "pressure".
     
  5. Mar 5, 2009 #4

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    This is not really a relativity question since neither time dilation nor length contraction are involved. Just classical Doppler effect.

    Let's say both mirrors are created at t=0 with a spatial separation x and stay stationary
    t = 0 : B see nothing yet
    t = x/c : B sees A appear
    t = 2x/c : B sees B appear in A
    t = 3x/c : B sees A appear in B which is already appearing in A
    and so forth...

    Now, what does relative inertial movement (B away from A) change? The duration between the appearance time points will be increasing. And the reflected inner pictures will be more and more red-shifted, so after a while you cannot see them with bare eyes. But in a finite time they will not be red-shifted to zero, so you can always detect a mirror appear in a black mirror, if you wait long enough.
     
  6. Mar 5, 2009 #5

    Ich

    User Avatar
    Science Advisor

    Hey, its a Gedankenexperiment. Given two perfect mirrors approaching at constant speed v, it mathematically follows that...
    In reality, of course, the photon will leave the gap rather sooner than later. And even if it didn't - theoretically - you couldn't keep the speed of the mirrors constant with finite force.
     
  7. Mar 5, 2009 #6

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    In reality of course there are other forces, acting on the mirrors than just the force of that one photon. At some point the electromagnetic forces prevent them coming closer, so the reflection rate will not get infinite. I am also not sure if the reflection of a photon is instantaneous.
     
  8. Mar 5, 2009 #7
    You don't even need a photon inbetween the mirrors. Suppose you have a box with reflecting walls. The box is empy, but this means that the electromagnetic field inside the box is in its ground state. The energy of the zero point energy in the box depends on the volume and this then leads to a pressure.

    If you compress the box slowly then the box will remain in its ground state (adiabatic approximation of quantum mechanics applies). If you compress it very fast then the quantum mechanical state vector will remain the same (Sudden Approximation). But the new vacuum of the compressed box is not the same state as the old vacuum. The old vacuum can be written as a superposition of the excited states of the electromagnetic field in the compressed box, so you will create photons.
     
  9. Mar 5, 2009 #8
    I think you're forgetting that these mirrors are observers. I thought of this question while studying minkowski diagrams and thinking about the zig-zag. Let's say that A's clock is t and B's clock is t'. then at t = 10, t' = 8.6 and at t' = 8.6, t = 7.5.
    The reason I asked is because I realized that this zig-zag apparently had no end.
    That's SR, right?

    speaking of which, why does this not break causality?
    Scenario:
    A and B start out 5 light-minutes from each other. observer B is traveling toward observer A at 0.5c. After 10 minutes of travel, A and B are at the same point. A is observing B as he is at t=8.6 and tells B (at t=8.6) some message. B then relays the message to A, who he's observing at t=7.5. The message says "send B this message in 2.5 seconds".
     
  10. Mar 5, 2009 #9
    Ah. that clears that up rather nicely (and has some interesting implications). Now to find a perfect mirror. hm... =]

    Thank you, everyone, for your informative replies. I've been wrestling with this topic for so long.
     
  11. Mar 5, 2009 #10

    JesseM

    User Avatar
    Science Advisor

    Where do you get that number? After 10 seconds at 0.5c in the mirror A frame, mirror B is 5 light-seconds away in this frame, so the light hitting B at that moment must have been emitted from mirror A 5 seconds earlier, at t=5 seconds in the mirror A frame. Perhaps you meant 10 seconds of B's own time? In this case it would actually be [tex]10/\sqrt{1 - 0.5^2}[/tex] = 11.547 seconds in the A frame, at which point B is 5.7735 light-seconds away in this frame, so the light B at that moment must have been emitted from mirror A 5.7735 seconds earlier at t=5.7735 seconds in A's frame.

    I guess you may have been thinking in terms of simultaneity, since if mirror B has coordinates x=5, t=10 in the A frame, and we use the Lorentz transformation to find the t' coordinate of this event in the B frame, we get t' = 1.1547 (10 - 0.5*5) = 8.66 which is close to what you got (maybe just some roundoff error). But mirrors don't show you what distant objects look like simultaneously in the mirror frame--that would imply the light from the object to the mirror would have to travel infinitely fast in the mirror's rest frame! It's important to understand that in relativity, the event of my seeing light from a distant event is not simultaneous with the event itself...instead, simultaneity is based on factoring out the travel time for the light to reach me, like if I see an event 10-light years away in my frame in 2009 according to my clock, then in my frame the event was actually simultaneous with the event of my clock reading 1999.
     
  12. Mar 5, 2009 #11
    That's exactly what I meant.
    I was thinking in terms of simultaneity. I made the exact mistake that you proposed. So, without doing calculations, I'll presume that the numbers work out to prevent causality (which would be broken if c was infinite... but I'm sure there's many examples proving that). Thank you very much; you saw my trip up exactly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook